Consider an infinite geometric series with first term a and common ratio r. if the sum is 4 and the sencond term is 3/4 ,then

Options

A. a=23 , ` r = 3/8`
B. `a = 4/7 , r = 3/7`
C. ` a = 3/2 , r = 1/2`
D. ` a = 3, r =1/4`

Answer

D

Explanation

Sum of infinite geometric series,
` S_(oo) = a/(1-r)`
` Rightarrow 4 = a/(1 -r) Rightarrow a = 4 -4r` …(i)
Also ` T_(2) = ar Rightarrow ar = 3/4` ….(ii)
From Eqs. (i) and (ii) , we get
` a = 4-4 (3/(4a))`
` a = 4- 3/a`
` Rightarrow a^(2) -4a+ 3=0`
` Rightarrow a^(2) -3a -a+3=0`
` Rightarrow a (a-3) -1 (a-3) =0`
`Rightarrow a=1,3`
On putting a=3 in Eq. (ii) ,we get
` 3r = 3/4`
` Rightarrow r =1/4`